Chứng minh rằng:
a) \(2\sin \left( {\frac{\pi }{4} + \alpha } \right)\sin \left( {\frac{\pi }{4} - \alpha } \right) = \cos 2\alpha \)
b) \(\sin \alpha \left( {1 + \cos 2\alpha } \right) = \sin 2\alpha \cos \alpha \)
c) \(\frac{{1 + \sin 2\alpha - \cos 2\alpha }}{{1 + \sin 2\alpha + \cos 2\alpha }} = \tan \alpha \) (khi các biểu thức có nghĩa)
d) \(\tan \alpha - \frac{1}{{\tan \alpha }} = - \frac{2}{{\tan 2\alpha }}\) (khi các biểu thức có nghĩa)
a)
\(\begin{array}{l}
2\sin \left( {\frac{\pi }{4} + \alpha } \right)\sin \left( {\frac{\pi }{4} - \alpha } \right)\\
= \cos 2\alpha - \cos \frac{\pi }{2} = \cos 2\alpha
\end{array}\)
b)
\(\begin{array}{*{20}{l}}
\begin{array}{l}
\sin \alpha \left( {1 + \cos 2\alpha } \right)\\
= \sin \alpha \left( {1 + 2{{\cos }^2}\alpha - 1} \right)
\end{array}\\
{ = 2\sin \alpha {{\cos }^2}\alpha = \sin 2\alpha \cos \alpha }
\end{array}\)
c)
\(\begin{array}{*{20}{l}}
\begin{array}{l}
\frac{{1 + \sin 2\alpha - \cos 2\alpha }}{{1 + \sin 2\alpha + \cos 2\alpha }}\\
= \frac{{\sin 2\alpha \left( {1 - \cos 2\alpha } \right)}}{{\sin 2\alpha \left( {1 + \cos 2\alpha } \right)}}
\end{array}\\
\begin{array}{l}
= \frac{{\sin 2\alpha + 2{{\sin }^2}\alpha }}{{\sin 2\alpha + 2{{\cos }^2}\alpha }}\\
= \frac{{2\sin \alpha \left( {\cos \alpha + \sin \alpha } \right)}}{{2\cos \alpha \left( {\cos \alpha + \sin \alpha } \right)}} = \tan \alpha
\end{array}
\end{array}\)
d)
\(\begin{array}{l}
\tan \alpha - \frac{1}{{\tan \alpha }} = 2.\frac{{{{\tan }^2}\alpha - 1}}{{2\tan \alpha }}\\
= - \frac{2}{{\tan 2\alpha }}
\end{array}\)
-- Mod Toán 10
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